How do you solve Search a 2D Matrix II on LeetCode?

Search a 2D Matrix II (LeetCode #240) can be solved using Brute Force or Optimal Solution. The optimal approach is Optimal Solution with O(n + m) time complexity and O(1) space complexity. Key insight: Utilizing the sorted properties of the matrix allows for more efficient searching.

What is the brute force solution for Search a 2D Matrix II?

The brute force approach for Search a 2D Matrix II is Brute Force, with O(n²) time complexity and O(1) space complexity. The simplest way to find the target is to check every element in the matrix. Since the matrix is sorted, we can iterate through each row and each column to see if the target exists. The optimal Optimal Solution approach improves this to O(n + m).

What algorithm or data structure is used to solve Search a 2D Matrix II?

Search a 2D Matrix II uses the following concepts: Array, Binary Search, Divide and Conquer, Matrix. The recommended patterns to study are: Two Pointers, Binary Search, Matrix Traversal.

What are the interview tips for Search a 2D Matrix II?

Always clarify the properties of the input data before jumping into coding. Think about edge cases, such as searching for the smallest or largest element. Practice explaining your thought process as you code; it helps in interviews.

What are common mistakes when solving Search a 2D Matrix II?

Not recognizing that we can eliminate entire rows or columns based on comparisons. Getting stuck in a loop without considering the bounds of the matrix.

#240

Search a 2D Matrix II

Medium

ArrayBinary SearchDivide and ConquerMatrixTwo PointersBinary SearchMatrix Traversal

LeetCode ↗

Approaches

Brute ForceOptimal

Complexity Comparison

	Brute Force	Optimal Solution★
Time	O(n²)	O(n + m)
Space	O(1)	O(1)

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Intuition

Time O(n + m)Space O(1)

We can leverage the sorted properties of the matrix. Start from the top-right corner and move left if the current number is greater than the target or down if it’s less than the target. This way, we can eliminate entire rows or columns in each step.

⚙️

Algorithm

6 steps

1Step 1: Start at the top-right corner of the matrix.
2Step 2: Compare the current element with the target.
3Step 3: If the current element equals the target, return true.
4Step 4: If the current element is greater than the target, move left (decrease column index).
5Step 5: If the current element is less than the target, move down (increase row index).
6Step 6: Repeat until you go out of bounds or find the target.

solution.py10 lines

1# Full working Python code
2row, col = 0, len(matrix[0]) - 1
3while row < len(matrix) and col >= 0:
4    if matrix[row][col] == target:
5        return True
6    elif matrix[row][col] > target:
7        col -= 1
8    else:
9        row += 1
10return False

ℹ

Complexity note: This complexity is efficient because we only traverse the matrix in a linear fashion, either moving left or down, which results in at most m + n steps.

1Utilizing the sorted properties of the matrix allows for more efficient searching.
2Moving strategically through the matrix can reduce the number of comparisons needed.

Solutions and explanations are original Tejav content. Problem titles © LeetCode — use the LeetCode button above for the full problem statement.